Question

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00i + 8.00j) m/s^2

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00i + 8.00j) m/s^2. It moves at constant speed. At time t2 = 3.00s (3/4 of a revolution later), it's acceleration is (8.00i – 9.00j) m/s^2. Find the radius of the path taken by the particle.

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